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RS Aggarwal Class 9 Solutions Chapter-3.pdf
- 1. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
FACTORISATION OF POLYNOMIALS - CHAPTER β 3
EXERCISE β 3A
Answer1. 9ππ
+ 12ππ
9π₯2
+ 12π₯π¦ = 3π₯(3π₯) + 3π₯(4π¦)
= 3π₯(3π₯ + 4π¦)
Answer2. 18ππ
π- 24πππ
18π₯2
π¦ - 24π₯π¦π§ = 6π₯π¦(3π₯) β 6π₯π¦(4π§)
= 6π₯π¦(3π₯ β 4π§)
Answer3. 27ππ
ππ
β ππππ
ππ
27π3
π3
β 45π4
π2
= 9π3
π2(3π) β 9π3
π2
(5π)
= 9π3
π2
(3π β 5π)
Answer4. ππ(π + π) β ππ(π + π)
2π(π₯ + π¦) β 3π(π₯ + π¦) = (π₯ + π¦)(2π β 3π)
Answer5. 2π(ππ
+ ππ) + ππ(ππ
+ ππ
)
2π₯(π2
+ π2) + 4π¦(π2
+ π2
) = (π2
+ π2)(2π₯ + 4π¦)
= 2(π2
+ π2)(π₯ + 2π¦)
Answer6. π(π β π) + π(π β π)
.π₯(π β 5) + π¦(π β 5) = (π β 5)(π₯ + π¦)
Answer7. 4(π + π) β π(π + π)π
4(π + π) β 6(π + π)2
= 2(π + π)(2 β 3(π + π))
= 2(π + π)(2 β 3π β 3π)
Answer8. 8(ππ β ππ)π
β ππ(ππ β ππ)
8(3π β 2π)2
β 10(3π β 2π) = 2(3π β 2π)(4(3π β 2π) β 5)
= 2(3π β 2π)(12π β 8π β 5)
Answer9. π(π + π)π
β πππ
π(π + π)
x(π₯ + π¦)3
β 3π₯2
π¦(π₯ + π¦) = π₯(π₯ + π¦)((π₯ + π¦)2
β 3π₯π¦)
= π₯(π₯ + π¦)(π₯2
+ π¦2
+ 2π₯π¦ β 3π₯π¦)
= π₯(π₯ + π¦)(π₯2
+ π¦2
β π₯π¦)
Answer10. ππ
+ πππ
+ ππ + ππ
π₯3
+ 2π₯2
+ 5π₯ + 10 = π₯3
+ 5π₯ + 2π₯2
+ 10
= π₯(π₯2
+ 5) + 2(π₯2
+ 5)
= (π₯ + 2)(π₯2
+ 5)
Answer11. ππ
+ ππ β πππ β πππ
.π₯2
+ π₯π¦ β 2π₯π§ β 2π¦π§ = π₯2
β 2π₯π§ + π₯π¦ β 2π¦π§
= π₯(π₯ β 2π§) + π¦(π₯ β 2π§)
= (π₯ + π¦)(π₯ β 2π§)
- 2. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer12. (ππ
π β ππ
π + πππ β ππ)
(π3
π β π2
π + 5ππ β 5π) = (π3
π + 5ππ β π2
π β 5π)
= [ππ(π2
+ 5) β π(π2
+ 5)]
= (ππ β π)(π2
+ 5)
= π(π β 1)(π2
+ 5)
Answer13. π β ππ β πππ
+ ππ
8 β 4π β 2π3
+ π4
= 8 β 2π3
β 4π + π4
= 2(4 β π3
) β π(4 + π3
)
= (2 β π)(4 β π3
)
Answer14. ππ
β πππ
π + ππππ
β πππ
.π₯3
β 2π₯2
π¦ + 3π₯π¦2
β 6π¦3
= π₯2(π₯ β 2π¦) + 3π¦2
(π₯ β 2π¦)
= (π₯ β 2π¦)(π₯2
+ 3π¦2
)
Answer15. ππ β ππ + ππ β ππ
pπ₯ β 5π + ππ β 5π₯ = ππ₯ β 5π₯ + ππ β 5π
= π₯(π β 5) + π(π β 5)
= (π₯ + π)(π β 5)
Answer16. ππ
+ π β ππ β π
.π₯2
+ π¦ β π₯π¦ β π₯ = π₯2
β π₯π¦ + π¦ β π₯
= π₯(π₯ β π¦) + (β1)(π₯ β π¦)
= (π₯ β 1)(π₯ β π¦)
Answer17. (ππ β π)π
β ππ + π
(3π β 1)2
β 6π + 2 = (3π)2
+ (1)2
β 2.3π. 1 β 6π + 2
= 9π2
+ 1 β 6π β 6π + 2
= 9π2
β 12π + 3
= 9π2
β 9π β 3π + 3
= 9π(π β 1) β 3(π β 1)
= (π β 1)(9π β 3)
= 3(π β 1)(3π β 1)
Answer18. (ππ β π)π
β ππ + ππ
(2π₯ β 3)2
β 8π₯ + 12 = (2π₯ β 3)2
β 4(2π₯ β 3)
= (2π₯ β 3)[(2π₯ β 3) β 4]
= (2π₯ β 3)(2π₯ β 7)
Answer19. ππ
+ π β πππ
β π
.π3
+ π β 3π2
β 3 = π(π2
+ 1) β 3(π2
+ 1)
= (π2
+ 1)(π β 3)
Answer20. πππ β πππ β πππ + πππ
3ππ₯ β 6ππ¦ β 8ππ¦ + 4ππ₯ = 3π(π₯ β 2π¦) + 4π(β2π¦ + π₯)
= (π₯ β 2π¦)(3π + 4π)
Answer21. ππππ
+ ππ
π + ππ
π + ππ
aππ₯2
+ π2
π₯ + π2
π₯ + ππ = ππ₯(ππ₯ + π) + π(ππ₯ + π)
= (ππ₯ + π)(ππ₯ + π)
- 3. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer22. ππ
β ππ
+ ππ + π β π β π
.π₯3
β π₯2
+ ππ₯ + π₯ β π β 1 = π₯3
β π₯2
+ π₯(π + 1) β 1(π + 1)
= π₯3
β π₯2
+ [(π₯ β 1)(π + 1)]
= π₯2(π₯ β 1) + [(π₯ β 1)(π + 1)]
= (π₯ β 1)[π₯2
+ (π + 1)]
= (π₯ β 1)(π₯2
+ π + 1)
Answer23. ππ + ππ β πππ β π
2π₯ + 4π¦ β 8π₯π¦ β 1 = 2π₯ β 1 + 4π¦ β 8π₯π¦
= 1(2π₯ β 1) β 4π¦[(β1) + 2π₯)]
= 1(2π₯ β 1) β 4π¦(2π₯ β 1)
= (2π₯ β 1)(1 β 4π¦)
Answer24. ππ(ππ
+ ππ) β ππ(ππ
+ ππ
)
aπ(π₯2
+ π¦2) β π₯π¦(π2
+ π2) = πππ₯2
+ πππ¦2
β π2
π₯π¦ β π2
π₯π¦
= πππ₯2
β π2
π₯π¦ + πππ¦2
β π2
π₯π¦
= ππ₯(ππ₯ β ππ¦) β ππ¦[(βππ¦) + ππ₯]
= ππ₯(ππ₯ β ππ¦) β ππ¦(ππ₯ β ππ¦)
= (ππ₯ β ππ¦)(ππ₯ β ππ¦)
Answer25. ππ
+ ππ(π + π) + ππ
.π2
+ ππ(π + 1) + π3
= π2
+ ππ2
+ ππ + π3
= π(π + π2) + π(π + π2
)
= (π + π)(π + π2
)
Answer26. ππ
+ ππ(π β ππ) β πππ
.π3
+ ππ(1 β 2π) β 2π2
= π3
+ ππ β 2π2
π β 2π2
= π(π2
+ π) β 2π(π2
+ π)
= (π β 2π)(π2
+ π)
Answer27. πππ
+ ππ β πππ β ππ
2π2
+ ππ β 2ππ β ππ = 2π2
β 2ππ + ππ β ππ
= 2π(π β π) β π(π β π)
= (π β π)(2π β π)
Answer28. (ππ + ππ)π
+ (ππ β ππ)π
(ππ₯ + ππ¦)2
+ (ππ₯ β ππ¦)2
= (ππ₯)2
+ (ππ¦)2
+ 2. ππ₯. ππ¦ + (ππ₯)2
+ (ππ¦)2
β 2. ππ₯. ππ¦
= π2
π₯2
+ π2
π¦2
+ π2
π₯2
+ π2
π¦2
+ (2πππ₯π¦ β 2πππ₯π¦)-
= π2
π₯2
+ π2
π¦2
+ π2
π₯2
+ π2
π¦2
= a2(π₯2
+ π¦2) + π2
(π₯2
+ π¦2
)
= (π₯2
+ π¦2
)(π2
+ π2
)
Answer29. π(π + π β π) β ππ
a(π + π β π) β ππ = π2
+ ππ β ππ β ππ
= π(π + π) β π(π + π)
= (π + π)(π β π)
Answer30. π(π β ππ β π) + πππ
a(π β 2π β π) + 2ππ = π2
β 2ππ β ππ + 2ππ
= π2
β ππ β 2ππ + 2ππ
= π(π β π) β 2π(π β π)
= (π β π)(π β 2π)
- 4. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer31. ππ
ππ
+ (πππ
+ π)π + π
.π2
π₯2
+ (ππ₯2
+ 1)π₯ + π = π2
π₯2
+ ππ₯3
+ π₯ + π
= ππ₯2(π + π₯) + 1(π + π₯)
= (π + π₯)(ππ₯2
+ 1)
Answer32. ππ(ππ
+ π) + π(ππ
+ ππ)
aπ(π₯2
+ 1) + π₯(π2
+ π2) = πππ₯2
+ ππ + π₯π2
+ π₯π2
= πππ₯2
+ π₯π2
+ π₯π2
+ ππ
= ππ₯(ππ₯ + π) + π(ππ₯ + π)
= (ππ₯ + π)(ππ₯ + π)
Answer33. ππ
β (π + π)π + ππ
.π₯2
β (π + π)π₯ + ππ = π₯2
β ππ₯ β ππ₯ + ππ
= π₯(π₯ β π) β π(π₯ β π)
= (π₯ β π)(π₯ β π)
Answer34. ππ
+
π
ππ β π β ππ +
π
π
.π₯2
+
1
π₯2 β 2 β 3π₯ +
3
π₯
= π₯2
+
1
π₯2 β 2. π₯.
1
π₯
β 3 (π₯ β
1
π₯
)
= (π₯ β
1
π₯
)
2
β 3 (π₯ β
1
π₯
)
= (π₯ β
1
π₯
) (π₯ β
1
π₯
β 3)
- 5. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
EXERCISE β 3B
FORMULA USED β: (ππ
β ππ) = (π β π)(π + π)
Answer.1. πππ
β ππππ
9π₯2
β 16π¦2
= (3π₯)2
β (4π¦)2
= (3π₯ β 4π¦)(3π₯ + 4π¦)
Answer.2. (
ππ
π
ππ
β
π
π
ππ)
.(
25
4
π₯2
β
1
9
π¦2) = (
25
4
π₯)
2
β (
1
9
π¦)
2
= (
25
4
π₯ β
1
9
π¦) (
25
4
π₯ +
1
9
π¦)
Answer.3. ππ β ππππ
81β16π₯2
= (9)2
β (4π₯)2
= (9 β 4π₯)(9 + 4π₯)
Answer.4. π β ππππ
5 β 20π₯2
= 5(1 β 4π₯2)
= 5[(1)2
β (2π₯)2]
= 5(1 β 2π₯)(1 + 2π₯)
Answer.5. πππ
β ππ
2π₯4
β 32 = 2(π₯4
β 16)
= 2[(π₯2)2
β (4)2
]
= 2(π₯2
β 4)(π₯2
+ 4)
= 2(π₯ β 2)(π₯ + 2)(π₯2
+ 4)
Answer.6. πππ
π β ππππππ
3π3
π β 243ππ3
= 3ππ(π2
β 81π2)
= 3ππ(π β 9π)(π + 9π)
Answer.7. πππ
β πππ
3π₯3
β 48π₯ = 3π₯(π₯2
β 16)
= 3π₯(π₯ β 4)(π₯ + 4)
Answer.8. ππππ
β ππππ
27π2
β 48π2
= 3(9π2
β 16π2)
= 3[(3π)2
β (4π)2
]
= 3(3π β 4π)(3π + 4π)
Answer.9. π β ππππ
.π₯ β 64π₯3
= π₯(1 β 64π₯2)
= π₯{(1)2
β (8π₯)2
}
= π₯(1 β 8π₯)(1 + 8π₯)
Answer.10. ππππ
β ππππ
8ππ2
β 18π3
= 2π(4π2
β 9π2)
= 2π(2π β 3π)(2π + 3π)
Answer.11. πππ β πππ
- 6. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
150 β 6π₯2
= 6(25 β π₯2)
= 6(5 β π₯)(5 + π₯)
Answer.12. π β ππππ
2 β 50π₯2
= 2(1 β 25π₯2)
= 2(1 β 5π₯)(1 + 5π₯)
Answer.13. ππππ
β ππ
20π₯2
β 45 = 5(4π₯2
β 9)
= 5(2π₯ β 3)(2π₯ + 3)
Answer.14. (ππ + ππ)π
β πππ
(3π + 5π)2
β 4π2
= (3π + 5π β 2π)(3π + 5π + 2π)
Answer.15. ππ
β ππ
β π β π
.π2
β π2
β π β π = (π β π)(π + π) β 1(π + π)
= (π + π)(π β π β 1)
Answer.16. πππ
β πππ
β ππ β ππ
4π2
β 9π2
β 2π β 3π = (2π β 3π)(2π + 3π) β 1(2π + 3π)
= (2π + 3π){(2π β 3π) β 1}
= (2π + 3π)(2π β 3π β 1)
Answer.17. ππ
β ππ
+ πππ β ππ
.π2
β π2
+ 2ππ β π2
= π2
β (π2
β 2ππ + π2)
= π2
β (π β π)2
= {π β (π β π)}{π + (π β π)}
= (π β π + π) (π + π β π)
Answer.18. πππ
β πππ
+ ππ + π
4π2
β 4π2
+ 4π + 1 = (4π2
+ 4π + 1) β 4π2
= (2π + 1)2
β (2π)2
= (2π + 1 + 2π)(2π + 1 β 2π)
= (2π + 2π + 1)(2π β 2π + 1)
Answer.19. ππ
+ πππ + ππ
β πππ
.π2
+ 2ππ + π2
β 9π2
= (π2
+ 2ππ + π2) β (3π)2
= (π + π)2
β (3π)2
= (π + π β 3π)(π + π + 3π)
Answer.20. πππππ
β π(π β π)π
108π2
β 3(π β π)2
= 3{36π2
β (π β π)2}
= 3{6π β (π β π)}{6π + (π β π)}
= 3(6π β π + π)(6π + π β π)
Answer.21. (π + π)π
β π β π
(π + π)3
β π β π = (π + π)3
β 1(π + π )
= (π + π){(π + π)2
β 1}
= (π + π)(π + π + 1)(π + π β 1)
Answer.22. ππ
+ ππ
β ππ
β πππ
.π₯2
+ π¦2
β π§2
β 2π₯π¦ = (π₯2
+ π¦2
β 2π₯π¦) β π§2
- 7. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (π₯ β π¦)2
β π§2
= (π₯ β π¦ β π§)(π₯ β π¦ + π§)
Answer.23. ππ
+ πππ + ππ
β ππ
+ πππ β ππ
.π₯2
+ 2π₯π¦ + π¦2
β π2
+ 2ππ β π2
= (π₯2
+ 2π₯π¦ + π¦2) β (π2
β 2ππ + π2)
= (π₯ + π¦)2
β (π β π)2
= {(π₯ + π¦) β (π β π)}{(π₯ + π¦) + (π β π)}
= π₯ + π¦ β π + π)(π₯ + π¦ + π β π)
Answer.24. ππππ
β πππ + π β ππππ
25π₯2
β 10π₯ + 1 β 36π¦2
= (25π₯2
β 10π₯ + 1) β (6π¦)2
= (5π₯ β 1)2
β (6π¦)2
= (5π₯ β 1 β 6π¦)(5π₯ β 1 + 6π¦)
Answer.25. π β π β ππ
+ ππ
a β π β π2
+ π2
= π β π β 1(π2
β π2)
= (π β π) β 1(π β π)(π + π)
= (π β π)(1 β π β π)
Answer.26. ππ
β ππ
β πππ + πππ
.π2
β π2
β 4ππ + 4π2
= (π2
β 4ππ + 4π2) β π2
= (π β 2π)2
β π2
= (π β 2π β π)(π β 2π + π)
Answer.27. π β ππ
+ πππ β ππ
9β π2
+ 2ππ β π2
= (3)2
β (π2
β 2ππ + π2)
= (3)2
β (π β π)2
= {3 β (π β π)}{3 + (π β π)}
= (3 β π + π)(3 + π β π)
Answer.28. ππ
β πππ
β π + π
.π₯3
β 5π₯2
β π₯ + 5 = π₯2(π₯ β 5) β 1(π₯ β 5)
= (π₯ β 5)(π₯2
β 1)
= (π₯ β 5)(π₯ β 1)(π₯ + 1)
Answer.29. π + πππ β (ππ
+ ππ)
1+2ππ β (π2
+ π2) = 1 β {(β2ππ} + (π2
+ π2)}
= 1 β (π β π)2
= {1 β (π β π)}{1 + (π β π)}
= (1 β π + π)(1 + π β π)
Answer.30. πππ
+ ππ + π β ππππ
9π2
+ 6π + 1 β 36π2
= (9π2
+ 6π + 1) β (6π)2
= (3π + 1)2
β (6π)2
= (3π + 1 β 6π)(3π + 1 + 6π)
Answer.31. ππ
βππ
+ ππ β π
.π₯2
βπ¦2
+ 6π¦ β 9 = π₯2
β (π¦2
β 6π¦ + 9)
= π₯2
β(π¦ β 3)2
= {π₯ β (π¦ β 3)}{π₯ + (π¦ β 3)}
= (π₯ β π¦ + 3)(π₯ + π¦ β 3)
- 8. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.32. πππ
β πππ
β ππ β ππ
4π₯2
β 9π¦2
β 2π₯ β 3π¦ = (2π₯)2
β (3π¦)2
β 1(2π₯ + 3π¦)
= (2π₯ β 3π¦)(2π₯ + 3π¦) β 1(2π₯ + 3π¦)
= (2π₯ + 3π¦)(2π₯ β 3π¦ β 1)
Answer.33. πππ
+ ππ β ππ β ππππ
9π2
+ 3π β 8π β 64π2
= (3π)2
β (8π)2
+ 3π β 8π
= (3π β 8π)(3π + 8π) + 1(3π β 8π)
= (3π β 8π)(3π + 8π + 1)
Answer.34. ππ
+
π
ππ β π
.π₯2
+
1
π₯2 β 3 = (π₯2
+
1
π₯2 β 2. π₯.
1
π₯
) β 1
= (π₯ +
1
π₯
)
2
β 12
= (π₯ +
1
π₯
β 1) (π₯ +
1
π₯
+ 1)
Answer.35. ππ
β π +
π
ππ β ππ
.π₯2
β 2 +
1
π₯2 β π¦2
= (π₯2
β 2 +
1
π₯2
) β (π¦)2
= (π₯ β
1
π₯
)
2
β (π¦)2
= (π₯ β
1
π₯
β π¦)(π₯ β
1
π₯
+ π¦)
Answer.36. ππ
+
π
ππ
.π₯4
+
4
π₯4 = π₯4
+
4
π₯4 + 2. π₯2
.
2
π₯2 β 2. π₯2
.
2
π₯2
= (π₯4
+
4
π₯4 + 2. π₯2
.
2
π₯2
) β 4
=(π₯2
+
2
π₯2
) β (2)2
= (π₯2
+
2
π₯2 β 2) (π₯2
+
2
π₯2 + 2)
Answer.37. ππ
β π
.π₯8
β 1 = (π₯4)2
β 12
= (π₯4
β 1)(π₯4
+ 1)
= (π₯2
β 1)(π₯2
+ 1)(π₯4
+ 1 + 2. π₯2
. 1 β 2. π₯2
. 1)
= (π₯ β 1)(π₯ + 1)(π₯2
+ 1) {(π₯2
+ 1)2
β (β2π₯)
2
}
= (π₯ β 1)(π₯ + 1)(π₯2
+ 1)(π₯2
+ 1 β β2π₯)(π₯2
+ 1 + β2π₯)
Answer.38. ππππ
β π
16π₯4
β 1 = (4π₯2
)2
β 12
= (4π₯2
β 1)(4π₯2
+ 1)
= (2π₯ β 1)(2π₯ + 1)(4π₯2
+ 1)
Answer.39. ππππ
β ππ
81π₯4
β π¦4
= (9π₯2)2
β (π¦2)2
= (9π₯2
β π¦2
)(9π₯2
+ π¦2
)
= (3π₯ β π¦)(3π₯ + π¦)(9π₯2
+ π¦2
)
Answer.40. ππ
β πππ
.π₯4
β 625 = (π₯2)2
β (25)2
= (π₯2
β 25)(π₯2
+ 25)
= (π₯ β 5)(π₯ + 5)(π₯2
+ 25)
- 10. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
EXERCISE β 3C
Answer.1. ππ
+ πππ + ππ
.π₯2
+ 11π₯ + 30 = π₯2
+ 5π₯ + 6π₯ + 30
= π₯(π₯ + 5) + 6(π₯ + 5)
= (π₯ + 5)(π₯ + 6)
Answer.2. ππ
+ πππ + ππ
.π₯2
+ 18π₯ + 32 = π₯2
+ 16π₯ + 2π₯ + 32
= π₯(π₯ + 16) + 2(π₯ + 16)
= (π₯ + 16)(π₯ + 2)
Answer.3. ππ
+ πππ β ππ
.π₯2
+ 20π₯ β 69 = π₯2
+ 23π₯ β 3π₯ β 69
= π₯(π₯ + 23) β 3(π₯ + 23)
= (π₯ + 23)(π₯ β 3)
Answer.4. ππ
+ πππ β πππ
.π₯2
+ 19π₯ β 150 = π₯2
+ 25π₯ β 6π₯ β 150
= π₯(π₯ + 25) β 6(π₯ + 25)
= (π₯ + 25)(π₯ β 6)
Answer.5. ππ
+ ππ β ππ
.π₯2
+ 7π₯ β 98 = π₯2
+ 14π₯ β 7π₯ β 98
= π₯(π₯ + 14) β 7(π₯ + 14)
= (π₯ + 14)(π₯ β 7)
Answer.6. ππ
+ πβππ β ππ
.π₯2
+ 2β3π₯ β 24 = π₯2
+ 4β3π₯ β 2β3π₯ β 24
= π₯(π₯ + 4β3) β 2β3(π₯ + 4β3)
= (π₯ + 4β3)(π₯ β 2β3)
Answer.7. ππ
β πππ + ππ
.π₯2
β 21π₯ + 90 = π₯2
β 15π₯ β 6π₯ + 90
= π₯(π₯ β 15) β 6(π₯ β 15)
= (π₯ β 15)(π₯ β 6)
Answer.8. ππ
β πππ + πππ
.π₯2
β 22π₯ + 120 = π₯2
β 12π₯ β 10π₯ + 120
= π₯(π₯ β 12) β 10(π₯ β 12)
= (π₯ β 12)(π₯ β 10)
Answer.9. ππ
β ππ + π
.π₯2
β 4π₯ + 3 = π₯2
β 3π₯ β π₯ + 3
= π₯(π₯ β 3) β 1(π₯ β 3)
= (π₯ β 3)(π₯ β 1)
Answer.10. ππ
+ πβππ + ππ
.π₯2
+ 7β6π₯ + 60 = π₯2
+ 5β6π₯ + 2β6π₯ + 60
= π₯(π₯ + 5β6) + 2β6(π₯ + 5β6)
= (π₯ + 5β6)(π₯ + 2β6)
Answer.11. ππ
+ πβππ + π
- 11. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
.π₯2
+ 3β3π₯ + 6 = π₯2
+ 2β3π₯ + β3π₯ + 6
= π₯(π₯ + 2β3) + β3(π₯ + 2β3)
= (π₯ + 2β3)(π₯ + β3)
Answer.12. ππ
+ πβππ + ππ
.π₯2
+ 6β6π₯ + 48 = π₯2
+ 4β6π₯ + 2β6π₯ + 48
= π₯(π₯ + 4β6) + 2β6(π₯ + 4β6)
= (π₯ + 4β6)(π₯ + 2β6)
Answer.13. ππ
+ πβππ + ππ
.π₯2
+ 5β5π₯ + 30 = π₯2
+ 3β5π₯ + 2β5π₯ + 30
= π₯(π₯ + 3β5) + 2β5(π₯ + 3β5)
= (π₯ + 3β5)(π₯ + 2β5)
Answer.14. ππ
β πππ β πππ
.π₯2
β 24π₯ β 180 = π₯2
β 30π₯ + 6π₯ β 180
= π₯(π₯ β 30) + 6(π₯ β 30)
= (π₯ β 30)(π₯ + 6)
Answer.15. ππ
β πππ β πππ
.π₯2
β 32π₯ β 105 = π₯2
β 35π₯ + 3π₯ β 105
= π₯(π₯ β 35) + 3(π₯ β 35)
= (π₯ β 35)(π₯ + 3)
Answer.16. ππ
β πππ β ππ
.π₯2
β 11π₯ β 80 = π₯2
β 16π₯ + 5π₯ β 80
= π₯(π₯ β 16) + 5(π₯ β 16)
= (π₯ β 16)(π₯ + 5)
Answer.17. π β π β ππ
.6 β π₯ β π₯2
= βπ₯2
β 3π₯ + 2π₯ + 6
= βπ₯(π₯ + 3) + 2(π₯ + 3)
= (βπ₯ + 2)(π₯ + 3)
= (2 β π₯)(π₯ + 3)
Answer.18. ππ
β βππ β π
.π₯2
β β3π₯ β 6 = π₯2
β 2β3π₯ + β3π₯ β 6
= π₯(π₯ β 2β3) + β3(π₯ β 2β3)
= (π₯ β 2β3)(π₯ + β3)
Answer.19. ππ + ππ β ππ
.40 + 3π₯ β π₯2
= βπ₯2
+ 8π₯ β 5π₯ + 40
=βπ₯(π₯ β 8) β 5(π₯ β 8)
= (π₯ β 8)(βπ₯ β 5)
= (8 β π₯)(π₯ + 5)
- 12. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.20. ππ
β πππ + πππ
.π₯2
β 26π₯ + 133 = π₯2
β 19π₯ β 7π₯ + 133
= π₯(π₯ β 19) β 7(π₯ β 19)
= (π₯ β 19)(π₯ β 7)
Answer.21. ππ
β πβππ β ππ
.π₯2
β 2β3π₯ β 24 = π₯2
β 4β3π₯ + 2β3π₯ β 24
= π₯(π₯ β 4β3) + 2β3(π₯ β 4β3)
= (π₯ β 4β3)(π₯ + 2β3)
Answer.22. ππ
β πβππ β ππ
.π₯2
β 3β5π₯ β 20 = π₯2
β 4β5π₯ + β5π₯ β 20
= π₯(π₯ β 4β5) + β5(π₯ β 4β5)
= (π₯ β 4β3)(π₯ + β5)
Answer.23. ππ
+ βππ β ππ
.π₯2
+ β2π₯ β 24 = π₯2
+ 4β2π₯ β 3β2π₯ β 24
= π₯(π₯ + 4β2) β 3β2(π₯ + 4β2)
= (π₯ + 4β2)(π₯ β 3β2)
Answer.24. ππ
β πβππ β ππ
.π₯2
β 2β2π₯ β 30 = π₯2
β 5β2π₯ + 3β2π₯ β 30
= π₯(π₯ β 5β2) + 3β2(π₯ β 5β2)
= (π₯ β 5β2)(π₯ + 3β2)
Answer.25. ππ
β π β πππ
.π₯2
β π₯ β 156 = π₯2
β 13π₯ + 12π₯ + 156
= π₯(π₯ β 13) + 12(π₯ β 13)
= (π₯ β 13)(π₯ + 12)
Answer.26. ππ
β πππ β πππ
.π₯2
β 32π₯ β 105 = π₯2
β 35π₯ + 3π₯ β 105
= π₯(π₯ β 35) + 3(π₯ β 35)
= (π₯ β 35)(π₯ + 3)
Answer.27. πππ
+ πππ + π
9π₯2
+ 18π₯ + 8 = 9π₯2
+ 12π₯ + 6π₯ + 8
= 3π₯(3π₯ + 4) + 2(3π₯ + 4)
= (3π₯ + 4)(3π₯ + 2)
Answer.28. πππ
+ πππ + ππ
6π₯2
+ 17π₯ + 12 = 6π₯2
+ 9π₯ + 8π₯ + 12
= 3π₯(2π₯ + 3) + 4(2π₯ + 3)
= (2π₯ + 3)(3π₯ + 4)
Answer.29. ππππ
+ ππ β ππ
18π₯2
+ 3π₯ β 10 = 18π₯2
+ 15π₯ β 12π₯ β 10
= 3π₯(6π₯ + 5) β 2(6π₯ + 5)
= (6π₯ + 5)(3π₯ β 2)
Answer.30. πππ
+ πππ β ππ
- 13. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
2π₯2
+ 11π₯ β 21 = 2π₯2
+ 14π₯ β 3π₯ β 21
= 2π₯(π₯ + 7) β 3(π₯ + 7)
= (π₯ + 7)(2π₯ β 3)
Answer.31. ππππ
+ ππ β π
15π₯2
+ 2π₯ β 8 = 15π₯2
+ 12π₯ β 10π₯ β 8
= 3π₯(5π₯ + 4) β 2(5π₯ + 4)
= (5π₯ + 4)(3π₯ β 2)
Answer.32. ππππ
+ ππ β π
21π₯2
+ 5π₯ β 6 = 21π₯2
+ 14π₯ β 9π₯ β 6
= 7π₯(3π₯ + 2) β 3(3π₯ + 2)
= (3π₯ + 2)(7π₯ β 3)
Answer.33. ππππ
β πππ + ππ
24π₯2
β 41π₯ + 12 = 24π₯2
β 32π₯ β 9π₯ + 12
= 8π₯(3π₯ β 4) β 3(3π₯ β 4)
= (3π₯ β 4)(8π₯ β 3)
Answer.34. πππ
β πππ + π
3π₯2
β 14π₯ + 8 = 3π₯2
β 12π₯ β 2π₯ + 8
= 3π₯(π₯ β 4) β 2(π₯ β 4)
= (π₯ β 4)(3π₯ β 2)
Answer.35. πππ
+ ππ β ππ
2π₯2
+ 3π₯ β 90 = 2π₯2
+ 15π₯ β 12π₯ β 90
= π₯(2π₯ + 15) β 6(2π₯ + 15)
= (2π₯ + 15)(π₯ β 6)
Answer.36. βπππ
+ ππ β πβπ
β5π₯2
+ 2π₯ β 3β5 = β5π₯2
+ 5π₯ β 3π₯ β 3β5
= β5π₯(π₯ + β5) β 3(π₯ + β5)
= (π₯ + β5)(β5π₯ β 3)
Answer.37. πβπππ
+ π β πβπ
2β3π₯2
+ π₯ β 5β3 = 2β3π₯2
+ 6π₯ β 5π₯ β 5β3
= 2β3π₯(π₯ + β3) β 5(π₯ + β3)
= (π₯ + β3)(2β3π₯ β 5)
Answer.38. πππ
+ πβπππ + π
7π₯2
+ 2β14π₯ + 2 = 7π₯2
+ β14π₯ + β14π₯ + 2
= β7π₯(β7π₯ + β2) + β2(β7π₯ + β2)
= (β7π₯ + β2)(β7π₯ + β2)
Answer.39. πβπππ
β πππ + πβπ
6β3π₯2
β 47π₯ + 5β3 = 6β3π₯2
β 45π₯ β 2π₯ + 5β3
= 3β3π₯(2π₯ β 5β3) β 1(2π₯ β 5β3)
= (2π₯ β 5β3)(3β3π₯ β 1)
Answer.40. πβπππ
+ πππ + πβπ
- 14. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
5β5π₯2
+ 20π₯ + 3β5 = 5β5π₯2
+ 15π₯ + 5π₯ + 3β5
= 5π₯(β5π₯ + 3) + β5(β5π₯ + 3)
= (β5π₯ + 3)(5π₯ + β5)
Answer.41. βπππ
+ πππ + πβπ
.β3π₯2
+ 10π₯ + 8β3 = β3π₯2
+ 6π₯ + 4π₯ + 8β3
= β3π₯(π₯ + 2β3) + 4(π₯ + 2β3)
= (π₯ + 2β3)(β3π₯ + 4)
Answer.42. βπππ
+ ππ + βπ
.β2π₯2
+ 3π₯ + β2 = β2π₯2
+ 2π₯ + π₯ + β2
= β2π₯(π₯ + β2) + 1(π₯ + β2)
= (π₯ + β2)(β2π₯ + 1)
Answer.43. πππ
+ πβππ + π
2π₯2
+ 3β3π₯ + 3 = 2π₯2
+ 2β3π₯ + β3π₯ + 3
= 2π₯(π₯ + β3) + β3(π₯ + β3)
= (π₯ + β3)(2π₯ + β3)
Answer.44. ππππ
β π β ππ
15π₯2
β π₯ β 28 = 15π₯2
β 21π₯ + 20π₯ β 28
= 3π₯(5π₯ β 7) + 4(5π₯ β 7)
= (5π₯ β 7)(3π₯ + 4)
Answer.45. πππ
β ππ β ππ
6π₯2
β 5π₯ β 21 = 6π₯2
β 14π₯ + 9π₯ β 21
= 2π₯(3π₯ β 7) + 3(3π₯ β 7)
= (3π₯ β 7)(2π₯ + 3)
Answer.46. πππ
β ππ β ππ
2π₯2
β 7π₯ β 15 = 2π₯2
β 10π₯ + 3π₯ β 15
= 2π₯(π₯ β 5) + 3(π₯ β 5)
= (π₯ β 5)(2π₯ + 3)
Answer.47. πππ
β πππ β ππ
5π₯2
β 16π₯ β 21 = 5π₯2
β 21π₯ + 5π₯ β 21
= π₯(5π₯ β 21) + 1(5π₯ β 21)
= (5π₯ β 21)(π₯ + 1)
Answer.48. πππ
β πππ β ππ
6π₯2
β 11π₯ β 35 = 6π₯2
β 21π₯ + 10π₯ β 35
= 3π₯(2π₯ β 7) + 5(2π₯ β 7)
= (2π₯ β 7)(3π₯ + 5)
Answer.49. πππ
β ππ β ππ
9π₯2
β 3π₯ β 20 = 9π₯2
β 15π₯ + 12π₯ β 20
= 3π₯(3π₯ β 5) + 4(3π₯ β 5)
= (3π₯ β 5)(3π₯ + 4)
Answer.50. ππππ
β ππ β π
10π₯2
β 9π₯ β 7 = 10π₯2
β 14π₯ + 5π₯ β 7
= 2π₯(5π₯ β 7) + 1(5π₯ β 7)
- 15. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (5π₯ β 7)(2π₯ + 1)
Answer.51. ππ
β ππ +
π
ππ
.π₯2
β 2π₯ +
7
16
=
16π₯2 β 32π₯ + 7
16
=
16π₯2 β 28π₯ β 4π₯ + 7
16
=
4π₯ ( 4π₯ β 7 )β 1 ( 4π₯ β 7 )
16
=
( 4π₯ β 7 )( 4π₯ β 1 )
16
= (
4π₯ β 7
16
) (4π₯ β 1)
= (
π₯
4
β
7
16
) (4π₯ β 1)
Answer.52.
π
π
ππ
β ππ β π
.
1
3
π₯2
β 2π₯ β 9 =
π₯2 β 6π₯ β 27
3
=
π₯2 β 9π₯ + 3π₯ β 27
3
=
π₯ ( π₯ β 9 ) + 3 ( π₯ β 9 )
3
=
( π₯ β 9 ) ( π₯ + 3 )
3
= (
π₯ + 3
3
) (π₯ β 9)
= (1 +
π₯
3
) (π₯ β 9)
Answer.53. ππ
+
ππ
ππ
π +
π
ππ
.π₯2
+
12
35
π₯ +
1
35
=
35π₯2 + 12π₯ + 1
35
=
35π₯2 + 7π₯ + 5π₯ + 1
35
=
7π₯ ( 5π₯+ 1 ) + 1 ( 5π₯+ 1 )
35
=
( 5π₯+ 1 ) ( 7π₯ + 1 )
35
= (5π₯ + 1) (
7π₯ + 1
35
)
= (5π₯ + 1) (
π₯
5
+
1
35
)
Answer.54. ππππ
β ππ +
π
ππ
21π₯2 β 2π₯ +
1
21
= 21π₯2 β π₯ β π₯ +
1
21
= 21π₯ (π₯ β
1
21
) β 1(π₯ β
1
21
)
= (π₯ β
1
21
) (21π₯ β 1)
Answer.55.
π
π
ππ
+ πππ + ππ
.
3
2
π₯2
+ 16π₯ + 10 =
3
2
π₯2
+ π₯ + 15π₯ + 10
=
π₯
2
(3π₯ + 2) + 5(3π₯ + 2)
= (3π₯ + 2) (
π₯
2
+ 5)
- 16. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.56.
π
π
ππ
β
ππ
π
π β ππ
.
2
3
π₯2 β
17
3
π₯ β 28 =
2π₯2
β 17π₯β 84
3
=
2π₯2 β 24π₯ +7π₯ β 84
3
=
2π₯ ( π₯ β 12 ) + 7 ( π₯ β 12 )
3
=
( π₯ β 12 ) ( 2π₯ +7 )
3
= (
π₯ β 12
3
)(2π₯ + 7)
= (
π₯
3
β
1
4
) (2π₯ + 7)
Answer.57.
π
π
ππ
β
ππ
π
π + π
.
3
5
π₯2 β
19
5
π₯ + 4 =
3π₯2
β 19π₯ + 20
5
=
3π₯2 β 15π₯ β 4π₯ + 20
5
=
3π₯ ( π₯ β 5 ) β 4 ( π₯ β 5 )
5
=
( π₯ β 5 ) ( 3π₯ β 4 )
5
= (
π₯ β 5
5
)(3π₯ β 4)
= (
π₯
5
β 1) (3π₯ β 4)
Answer.58. πππ
β π +
π
π
2π₯2
β π₯ +
1
8
=
16π₯2 β 8π₯ + 1
8
=
16π₯2 β 4π₯ β 4π₯ + 1
8
=
4π₯ ( 4π₯ β 1 ) β 1( 4π₯ β 1)
8
= (
4π₯ β 1
8
)(4π₯ β 1)
= (
π₯
2
β
1
8
) (4π₯ β 1)
Answer.59. π(π + π)π
β π(π + π) β π
Lππ‘ (π₯ + π¦) = π
2(π₯ + π¦)2
β 9(π₯ + π¦) β 5 = 2π2
β 9π β 5
= 2π2
β 10π + π β 5
= 2π(π β 5) + 1(π β 5)
= (π β 5)(2π + 1)
βΈ« 2(π₯ + π¦)2
β 9(π₯ + π¦) β 5 = (π₯ + π¦ β 5){2(π₯ + π¦) + 1} [β΅ π = (π₯ + π¦) ]
= (π₯ + π¦ β 5)(2π₯ + 2π¦ + 1)
Answer.60. π(ππ β π)π
β π(ππ β π) β ππ
Lππ‘ (2π β π) = π
9(2π β π)2
β 4(2π β π) β 13 = 9π2
β 4π β 13
= 9π2
β 13π + 9π β 13
= π(9π β 13) + 1(9π β 13)
= (9π β 13)(π + 1)
βΈ« 9(2π β π)2
β 4(2π β π) β 13 = {9(2π β π) β 13}{(2π β π) + 1} [β΅ π = (2π β π) ]
= (18π β 9π β 13)(2π β π + 1)
Answer.61. π(π β ππ)π
β ππ(π β ππ) + ππ
- 17. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Lππ‘ (π₯ β 2π¦) = π
7(π₯ β 2π¦)2
β 25(π₯ β 2π¦) + 12 = 7π2
β 25π + 12
= 7π2
β 21π β 4π + 12
= 7π(π β 3) β 4(π β 3)
= (π β 3)(7π β 4)
βΈ« 7(π₯ β 2π¦)2
β 25(π₯ β 2π¦) + 12 = (π₯ β 2π¦ β 3){7(π₯ β 2π¦) β 4} [β΅ π = (π₯ β 2π¦) ]
= (π₯ β 2π¦ β 3)(7π₯ β 14π¦ β 4)
Answer.62. ππ (ππ +
π
π
)
π
β (ππ +
π
π
) β π
Lππ‘ (3π₯ +
1
π₯
) = π
10 (3π₯ +
1
π₯
)
2
β (3π₯ +
1
π₯
) β 3 = 10π2
β π β 3
= 10π2
β 6π + 5π β 3
= 2π(5π β 3) + 1(5π β 3)
= (5π β 3)(2π + 1)
βΈ« 10 (3π₯ +
1
π₯
)
2
β (3π₯ +
1
π₯
) β 3 = {5 (3π₯ +
1
π₯
) β 3}{2 (3π₯ +
1
π₯
) + 1}[β΅ π = (3π₯ +
1
π₯
)]
= (15π₯ +
5
π₯
β 3) (6π₯ +
2
π₯
+ 1)
Answer.63. π (ππ β
π
π
)
π
+ π(ππ β
π
π
) β ππ
Lππ‘ (2π₯ β
3
π₯
) = π
6(2π₯ β
3
π₯
)
2
+ 7 (2π₯ β
3
π₯
) β 20 = 6π2
+ 7π β 20
= 6π2
+ 15π β 8π β 20
= 3π(2π + 5) β 4(2π + 5)
= (2π + 5)(3π β 4)
βΈ« 6 (2π₯ β
3
π₯
)
2
+ 7 (2π₯ β
3
π₯
) β 20 = {2 (2π₯ β
3
π₯
) + 5}{3 (2π₯ β
3
π₯
) β 4}[β΅ π = (2π₯ β
3
π₯
)]
= (4π₯ β
6
π₯
+ 5) (6π₯ β
9
π₯
β 4)
Answer.64. (π + ππ)π
+ πππ(π + ππ) + πππ
Lππ‘ (π + 2π) = π
.(π + 2π)2
+ 101(π + 2π) + 100 = π2
+ 101π + 100
= π2
+ 100π + π + 100
= π(π + 100) + 1(π + 100)
= (π + 100)(π + 1)
βΈ« (π + 2π)2
+ 101(π + 2π) + 100 = {(π + 2π) + 100}{(π + 2π) + 1} [β΅ π = (π + 2π) ]
= (π + 2π + 100)(π + 2π + 1)
Answer.65. πππ
+ πππ
β π
Lππ‘ π₯2
= π¦
4π₯4
+ 7π₯2
β 2 = 4π¦2
+ 7π¦ β 2
= 4π¦2
+ 8π¦ β π¦ β 2
= 4π¦(π¦ + 2) β 1(π¦ + 2)
= (π¦ + 2)(4π¦ β 1)
βΈ« 4π₯4
+ 7π₯2
β 2 = (π₯2
+ 2)(4π₯2
β 1)
= (π₯2
+ 2){(2π₯)2
β 12}
= (π₯2
+ 2)(2π₯ β 1)(2π₯ + 1)
- 18. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.66. {(πππ)π
β π}
.{(999)2
β 1} = {(999)2
β (1)2
}
= (999 β 1)(999 + 1)
= 998 Γ 1000
= 998000
- 19. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
EXERCISE β 3D
Formula Used -: (π + π + π)π
= ππ
+ ππ
+ ππ
+ πππ + πππ + πππ
Answer.1.
(i) (π + ππ + ππ)π
.(π + 2π + 5π)2
= (π)2
+ (2π)2
+ (5π)2
+ 2(π)(2π) + 2(2π)(5π) + 2(5π)(π)
= π2
+ 4π2
+ 25π2
+ 4ππ + 20ππ + 10ππ
(ii)(ππ β π + π)π
.(2π β π + π)2
= (2π)2
+ (βπ)2
+ (π)2
+ 2(2π)(βπ) + 2(βπ)(π) + 2(π)(2π)
= 4π2
+ π2
+ π2
β 4ππ β 2ππ + 4ππ
(iii) (π β ππ β ππ)π
.(π β 2π β 3π)2
= (π)2
+ (β2π)2
+ (β3π)2
+ 2(π)(β2π) + 2(β2π)(β3π) + 2(β3π)(π)
= π2
+ 4π2
+ 9π2
β 4ππ + 12ππ β 6ππ
Answer.2.
(i) (ππ β ππ β ππ)π
.(2π β 5π β 7π)2
= (2π)2
+ (β5π)2
+ (β7π)2
+ 2(2π)(β5π) + 2(β5π)(β7π) + 2(β7π)(2π)
= 4π2
+ 25π2
+ 49π2
β 20ππ + 70ππ β 28ππ
(ii) (βππ + ππ β ππ)π
.(β3π + 4π β 5π)2
= (β3π)2
+ (4π)2
+ (β5π)2
+ 2(β3π)(4π) + 2(4π)(β5π) + 2(β5π)(β3π)
= 9π2
+ 16π2
+ 25π2
β 24ππ β 40ππ + 30ππ
(iii)(
π
π
π β
π
π
π + π)
π
.(
1
2
π β
1
4
π + 2)
2
= (
1
2
π)
2
+ (β
1
4
π)
2
+ (2)2
+ 2 (
1
2
π) (β
1
4
π) + 2 (β
1
4
π) (2) + 2(2) (
1
2
π)
=
π2
4
+
π2
16
+ 4 β
ππ
4
β π + 2π
Answer.3. πππ
+ πππ
+ ππππ
+ ππππ β ππππ β ππππ
4π₯2
+ 9π¦2
+ 16π§2
+ 12π₯π¦ β 24π¦π§ β 16π§π₯ = (2π₯)2
+ (3π¦)2
+ (β4π§)2
+ {2(2π₯)(3π¦)} +
{2(3π¦)(β4π§)} + {2(β4π§)(2π₯)}
= {(2π₯) + (3π¦) + (β4π§)}2
= (2π₯ + 3π¦ β 4π§)2
= (2π₯ + 3π¦ β 4π§)(2π₯ + 3π¦ β 4π§)
Answer.4. πππ
+ ππππ
+ πππ
β ππππ + ππππ β ππππ
9π₯2
+ 16π¦2
+ 4π§2
β 24π₯π¦ + 16π¦π§ β 12π§π₯ = (β3π₯)2
+ (4π¦)2
+ (2π§)2
+ {2(β3π₯)(4π¦)} +
{2(4π¦)(2π§)} + {2(2π§)(β3π₯)}
= {(β3π₯) + (4π¦) + (2π§)}2
= (β3π₯ + 4π¦ + 2π§)2
= (β3π₯ + 4π¦ + 2π§)(β3π₯ + 4π¦ + 2π§)
Answer.5. ππππ
+ πππ
+ πππ
β ππππ β ππππ + ππππ
25π₯2
+ 4π¦2
+ 9π§2
β 20π₯π¦ β 12π¦π§ + 30π§π₯ = (5π₯)2
+ (β2π¦)2
+ (3π§)2
+ {2(5π₯)(β2π¦)} +
{2(β2π¦)(3π§)} + {2(3π§)(5π₯)}
= {(5π₯) + (β2π¦) + (3π§)}2
= (5π₯ β 2π¦ + 3π§)2
= (5π₯ β 2π¦ + 3π§)(5π₯ β 2π¦ + 3π§)
Answer.6. ππππ
+ πππ
+ πππ
β ππππ β ππππ + ππππ
16π₯2
+ 4π¦2
+ 9π§2
β 16π₯π¦ β 12π¦π§ + 24π§π₯ = (4π₯)2
+ (β2π¦)2
+ (3π§)2
+ {2(4π₯)(β2π¦)} +
{2(β2π¦)(3π§)} + {2(3π§)(4π₯)}
= {(4π₯) + (β2π¦) + (3π§)}2
= (4π₯ β 2π¦ + 3π§)2
- 20. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (4π₯ β 2π¦ + 3π§)(4π₯ β 2π¦ + 3π§)
Answer.7. (i) (ππ)π
(99)2
= (100 β 1)2
= (100)2
+ (1)2
β 2 Γ 100 Γ 1
= 10000 + 1 β 200
= 9801
(ii) (πππ)π
(995)2
= (1000 β 5)2
= (1000)2
+ (5)2
β 2 Γ 1000 Γ 5
= 1000000 + 25 β 10000
= 990025
(iii) (πππ)π
(107)2
= (100 + 7)2
= (100)2
+ (7)2
+ 2 Γ 100 Γ 7
= 10000 + 49 + 1400
= 11449
- 21. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
EXERCISE -3E
Formulae Used: (i) (π + π)π
= ππ
+ ππ
+ πππ(π + π)
(ii)(π β π)π
= ππ
β ππ
β πππ(π β π)
Answer.1.
(i) (ππ + π)π
.(3π₯ + 2)3
= (3π₯)3
+ (2)3
+ 3(3π₯)(2)(3π₯ + 2)
= 27π₯3
+ 8 + 18π₯(3π₯) + 18π₯(2)
= 27π₯3
+ 8 + 54π₯2
+ 36π₯
(ii) (ππ +
π
ππ
)
π
.(3π +
1
4π
)
3
= (3π)3
+ (
1
4π
)
3
+ 3(3π) (
1
4π
) (3π +
1
4π
)
= 27π3
+
1
64π3 +
9π
4π
(3π) +
9π
4π
(
1
4π
)
= 27π3
+
1
64π3 +
27π2
4π
+
9π
16π2
(iii) (π +
π
π
π)
π
.(1 +
2
3
π)
3
= (1)3
+ (
2
3
π)
3
+ 3(1) (
2
3
π) (1 +
2
3
π)
= 13
+
8
27
π3
+ 2π(1) + 2π (
2
3
π)
= 1 +
8
27
π3
+ 2π +
4
3
π2
= 1 +
8
27
π3
+
4
3
π2
+ 2π
Answer.2.
(i)(ππ β ππ)π
(5π β 3π)3
= (5π)3
β (3π)3
β 3(5π)(3π)(5π β 3π)
= 125π3
β 27π3
β 45ππ(5π) + 45ππ(3π)
= 125π3
β 27π3
β 225π2
π + 135ππ2
(ii)(ππ β
π
π
)
π
.(3π₯ β
5
π₯
)
3
= (3π₯)3
β (
5
π₯
)
3
β 3(3π₯) (
5
π₯
) (3π₯ β
5
π₯
)
= 27π₯3
β
125
π₯3 β 45(3π₯) + 45 (
5
π₯
)
= 27π₯3
β
125
π₯3 β 135π₯ +
225
π₯
(iii)(
π
π
π β π)
π
.(
4
5
π β 2)
3
= (
4
5
π)
3
β (2)3
β 3 (
4
5
π) (2) (
4
5
π β 2)
=
64
125
π3
β 8 β
24
5
π (
4
5
π) +
24
5
π(2)
=
64
125
π3
β 8 β
96
25
π2
+
48
5
π
Answer.3. πππ
+ ππππ
+ ππππ
π + πππππ
8π3
+ 27π3
+ 36π2
π + 54ππ2
= (2π)3
+ (3π)3
+ 3(2π)(3π)(2π) + 3(2π)(3π)(3π)
- 22. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (2π)3
+ (3π)3
+ 3(2π)(3π)(2π + 3π)
= (2π + 3π)3
= (2π + 3π)(2π + 3π)(2π + 3π)
Answer.4. ππππ
β ππππ
β πππππ
π + ππππππ
64π3
β 27π3
β 144π2
π + 108ππ2
= (4π)3
β (3π)3
β 3(4π)(3π)(4π) + 3(4π)(3π)(3π)
= (4π)3
β (3π)3
β 3(4π)(3π)(4π β 3π)
= (4π β 3π)3
= (4π β 3π)(4π β 3π)(4π β 3π)
Answer.5. π +
ππ
πππ
ππ
+
ππ
π
+
ππππ
ππ
1 +
27
125
π3
+
9π
5
+
27π2
25
= (1)3
+ (
3
5
π)
3
+ 3(1) (
3
5
π) (1) + 3(1) (
3
5
π) (
3
5
π)
= (1)3
+ (
3
5
π)
3
+ 3(1) (
3
5
π) (1 +
3
5
π)
= (1 +
3
5
π)
3
= (1 +
3
5
π) (1 +
3
5
π) (1 +
3
5
π)
Answer.6. πππππ
β ππππ
β πππππ
π + ππππππ
125π₯3
β 27π¦3
β 225π₯2
π¦ + 135π₯π¦2
= (5π₯)3
β (3π¦)3
β 3(5π₯)(3π¦)(5π₯) + 3(5π₯)(3π¦)(3π¦)
= (5π₯)3
β (3π¦)3
β 3(5π₯)(3π¦)(5π₯ β 3π¦)
= (5π₯ β 3π¦)3
= (5π₯ β 3π¦)(5π₯ β 3π¦)(5π₯ β 3π¦)
Answer.7. ππ
ππ
β πππ
πππ
+ ππππ
π β ππ
.π3
π₯3
β 3π2
ππ₯2
+ 3ππ2
π₯ β π3
= π3
π₯3
β π3
β 3π2
ππ₯2
+ 3ππ2
π₯
= (ππ₯)3
β (π)3
β 3(ππ₯)(π)(ππ₯) + 3(ππ₯)(π)(π)
= (ππ₯)3
β (π)3
β 3(ππ₯)(π)(ππ₯ β π)
= (ππ₯ β π)3
= (ππ₯ β π)(ππ₯ β π)(ππ₯ β π)
Answer.8.
ππ
πππ
ππ
β
ππ
ππ
ππ
+
ππ
π
π β π
.
64
125
π3
β
96
25
π2
+
48
5
π β 8 =
64
125
π3
β 8 β
96
25
π2
+
48
5
π
= (
4
5
π)
3
β (2)3
β 3 (
4
5
π) (2) (
4
5
π) + 3 (
4
5
π) (2)(2)
= (
4
5
π)
3
β (2)3
β 3 (
4
5
π) (2) (
4
5
π β 2)
= (
4
5
π β 2)
3
= (
4
5
π β 2) (
4
5
π β 2) (
4
5
π β 2)
Answer.9. ππ
β πππ(π β π) β ππ
- 23. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
.π3
β 12π(π β 4) β 64 = (π)3
β (4)3
β 3(π)(4)(π β 4)
= (π β 4)3
= (π β 4)(π β 4)(π β 4)
Answer.10.
(i) (πππ)π
(103)3
= (100 + 3)3
= (100)3
+ (3)3
+ 3 Γ 100 Γ 3 Γ (100 + 3)
= 1000000 + 27 + (900 Γ 103)
= 1000027 + 92700
= 1092727
(ii) (ππ)π
(99)3
= (100 β 1)3
= (100)3
β (1)3
β 3 Γ 100 Γ 1 Γ (100 β 1)
= 1000000 β 1 β (300 Γ 99)
= 999999 β 29700
= 970299
- 24. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
EXERCISE β 3F
Formulae Used: (i) (ππ
+ ππ) = (π + π)(ππ
β ππ + ππ)
(ii) (ππ
β ππ) = (π β π)(ππ
+ ππ + ππ
)
Answer.1. ππ
+ ππ
.π₯3
+ 27 = (π₯)3
+ (3)3
= (π₯ + 3)(π₯2
β 3π₯ + 9)
Answer.2. ππππ
+ ππππ
.27π3
+ 64π3
= (3π)3
+ (4π)3
=(3π + 4π)(9π2
β 12ππ + 16π2
)
Answer.3. πππππ
+
π
π
.125π3
+
1
8
= (5π)3
+ (
1
8
)
3
= (5π +
1
8
) (25π2
β
5π
2
+
1
4
)
Answer.4. πππππ
+
π
πππ
.216π₯3
+
1
125
= (6π₯)3
+ (
1
5
)
3
= (6π₯ +
1
5
) (36π₯2
β
6π₯
5
+
1
25
)
Answer.5. ππππ
+ πππ
.16π₯4
+ 54π₯ = 2π₯(8π₯3
+ 27)
= 2π₯{(2π₯)3
+ (3)3
}
= 2π₯(2π₯ + 3)(4π₯2
β 6π₯ + 6)
Answer.6. πππ
+ ππππ
.7π3
+ 56π3
= 7(π3
+ 8π3)
= 7{(π)3
+ (2π)3
}
= 7(π + 2π)(π2
β 2ππ + 4π2
)
Answer.7. ππ
+ ππ
.π₯5
+ π₯2
= π₯2
(π₯3
+ 1)
= π₯2
{(π₯)3
+ (1)3
}
= π₯2
(π₯ + 1)(π₯2
β π₯ + 1)
Answer.8. ππ
+ π. πππ
.π3
+ 0.008 = (π)3
+ (0.2)3
= (π + 0.2)(π2
β 0.2π + 0.04)
Answer.9. π β ππππ
.1 β 27π3
= (1)3
β (3π)3
= (1 β 3π)(1 + 3π + 9π2
)
Answer.10. ππππ
β πππ
.64π3
β 343 = (4π)3
β (7)3
= (4π β 7)(16π2
+ 28π + 49)
- 25. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.11. ππ
β πππ
.π₯3
β 512 = (π₯)3
β (8)3
= (π₯ β 8)(π₯2
+ 8π₯ + 64)
Answer.12. ππ
β π. πππ
.π3
β 0.064 = (π)3
β (0.4)3
= (π β 0.4)(π2
+ 0.4π + 0.16)
Answer.13. πππ
β
π
ππππ
.8π₯3
β
1
27π¦3 = (2π₯)3
β (
1
3π¦
)
3
= (2π₯ β
1
3π¦
) (4π₯2
+
2π₯
3π¦
+
1
9π¦2
)
Answer.14.
ππ
πππ
β πππ
.
π₯3
216
β 8π¦3
= (
π₯
6
)
3
β (2π¦)3
= (
π₯
6
+ 2π¦) (
π₯2
36
+
2π₯π¦
6
+ 4π¦2)
= (
π₯
6
+ 2π¦) (
π₯2
36
+
π₯π¦
3
+ 4π¦2)
Answer.15. π β ππππ
.π₯ β 8π₯π¦3
= π₯(1 β 8π¦3
)
= π₯{(1)3
β (2π¦)3
}
= π₯(1 β 2π¦)(1 + 2π¦ + 4π¦2
)
Answer.16. ππππ
β ππππ
.32π₯4
β 500π₯ = 4π₯(8π₯3
β 125)
= 4π₯{(2π₯)3
β (5)3
}
= 4π₯(2π₯ β 5)(4π₯2
+ 10π₯ + 25)
Answer.17. πππ
π β ππππ
ππ
3π7
π β 81π4
π4
= 3π4
π(π3
β 27π3
)
= 3π4
π{(π)3
β (3π)3
}
= 3π4
π(π β 3π)(π2
+ 3ππ + 9π2
)
Answer.18. ππ
ππ
β ππ
.π₯4
π¦4
β π₯π¦ = π₯π¦(π₯3
π¦3
β 1)
= π₯π¦{(π₯π¦)3
β (1)3
}
= π₯π¦(π₯π¦ β 1)(π₯2
π¦2
+ π₯π¦ + 1)
Answer.19. πππ
ππ
β ππ
8π₯2
π¦3
β π₯5
= π₯2(8π¦3
β π₯3)
= π₯2
{(2π¦)3
β (π₯)3
}
= π₯2
(2π¦ β π₯)(4π¦2
+ 2π₯π¦ + π₯2
)
Answer.20. ππππ β πππ
.1029 β 3π₯3
= 3(343 β π₯3
)
= 3{(7)3
β (π₯)3}
- 26. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= 3(7 β π₯)(49 + 7π₯ + π₯2
)
Answer.21. ππ
β πππ
.π₯6
β 729 = (π₯2
)3
β (9)3
= (π₯2
β 9){(π₯2)2
+ 9π₯2
+ 81)}
= (π₯ β 3)(π₯ + 3)(π₯4
+ 9π₯2
+ 81)
= (π₯ β 3)(π₯ + 3){(π₯4
+ 9π₯2
+ 9π₯2
+ 81) β (3π₯)2}
= (π₯ β 3)(π₯ + 3){(π₯2
+ 9)2
β (3π₯)2}
= (π₯ β 3)(π₯ + 3)(π₯2
+ 3π₯ + 9)(π₯2
β 3π₯ + 9)
Answer.22. ππ
β ππ
.π₯9
β π¦9
= (π₯3)3
β (π¦3)3
= (π₯3
β π¦3){(π₯3)2
+ π₯3
π¦3
+ (π¦3)2}
= {(π₯)3
β (π¦)3
}(π₯6
+ π₯3
π¦3
+ π¦6
)
= (π₯ β π¦)(π₯2
+ π₯π¦ + π¦2
)(π₯6
+ π₯3
π¦3
+ π¦6
)
Answer.23. (π + π)π
β (π β π)π
Lππ‘ (π + π) = π₯ πππ (π β π) = π¦
.(π₯3
β π¦3) = (π₯ β π¦)(π₯2
+ π₯π¦ + π¦2
)
Nππ€, ππ’π‘π‘πππ π£πππ’π ππ π₯ πππ π¦
. π
π»π = (π₯ β π¦)(π₯2
+ π₯π¦ + π¦2
)
= {(π + π) β (π β π)}{(π + π)2
+ (π + π)(π β π) + (π β π)2
}
= (π + π β π + π){π2
+ π2
+ 2ππ + π2
β ππ + ππ β π2
+ π2
+ π2
β 2ππ}
= (2π)(3π2
+ π2
)
Sπ,
.(π + π)3
β (π β π)3
= 2π(3π2
+ π2
)
Answer.24. πππ
β ππ
β πππ + πππ
8π3
β π3
β 4ππ₯ + 2ππ₯ = (2π)3
β (π)3
β 2π₯(2π β π)
= (2π β π)(4π2
+ 2ππ + π2) β 2π₯(2π β π)
= (2π β π)(4π2
+ 2ππ + π2
β 2π₯)
Answer.25. ππ
+ πππ
π + ππππ
+ ππ
β π
.π3
+ 3π2
π + 3ππ2
+ π3
β 8 = {π3
+ π3
+ 3ππ(π + π)} β (2)3
= (π + π)3
β (2)3
= (π + π β 2)[(π + π)2
+ 2(π + π) + 4]
Answer.26. ππ
β
π
ππ β ππ +
π
π
.π3
β
1
π3 β 2π +
2
π
= (π3
β
1
π3
) β (2π +
2
π
)
= (π β
1
π
) (π2
+ 1 +
1
π2
) β 2 (π β
1
π
)
= (π β
1
π
) (π2
+ 1 +
1
π2 β 2)
= (π β
1
π
) (π2
β 1 +
1
π2
)
Answer.27. πππ
+ ππππ
β ππ β πππ
2π3
+ 16π3
β 5π β 10π = 2(π3
+ 8π3
) β 5(π + 2π)
= 2(π + 2π)(π2
β 2ππ + 4π2) β 5(π + 2π)
= (π + 2π){2(π2
β 2ππ + 4π2) β 5}
= (π + 2π)(2π2
β 4ππ + 8π2
β 5)
Answer.28. ππ
+ ππ
.π6
+ π6
= (π2)3
+ (π2)3
- 27. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (π2
+ π2
){(π2)2
β π2
π2
+ (π2)2
}
= (π2
+ π2
)(π4
β π2
π2
+ π4
)
Answer.29. πππ
β πππ
.π12
β π12
= (π6)2
β (π6)2
= (π6
β π6)(π6
+ π6)
= [(π3)2
β(π3)2][(π2)3
+ (π2)3]
= (π3
β π3)(π3
+ π3)[(π2
+ π2
)(π4
β π2
π2
+ π4
)]
= (π β π)(π2
+ ππ + π2
)(π + π)(π2
β ππ + π2
)(π2
+ π2
)(π4
β π2
π2
+ π4
)
= (π β π)(π + π)(π2
+ π2)(π2
+ ππ + π2)(π2
β ππ + π2)(π4
β π2
π2
+ π4)
Answer.30. ππ
β πππ
β π
Lππ‘ π₯3
= π
.π₯6
β 7π₯3
β 8 = π2
β 7π β 8
= π2
β 8π + π β 8
= π(π β 8) + 1(π β 8)
= (π + 1)(π β 8)
Sπ, π₯6
β 7π₯3
β 8 = (π + 1)(π β 8)
= (π₯3
+ 1)(π₯3
β 8) [β΅ π = π₯3
]
= [(π₯)3
+ (1)3][(π₯)3
β (2)3
]
= (π₯ + 1)(π₯2
β π₯ + 1)(π₯ β 2)(π₯2
+ π₯ + 4)
= (π₯ β 2)(π₯ + 1)(π₯2
β π₯ + 1)(π₯2
+ π₯ + 4)
Answer.31. ππ
β πππ
+ ππ + π
.π₯3
β 3π₯2
+ 3π₯ + 7 = (π₯3
β 3π₯2
+ 3π₯ β 1) + 8
= [π₯3
β 13
β 3π₯(π₯ β 1)] + (2)3
= (π₯ β 1)3
+ (2)3
= [(π₯ β 1) + 2][(π₯ β 1)2
β (π₯ β 1) Γ (2) + (2)2
]
= (π₯ + 1)[(π₯2
β 2π₯ + 1) β 2π₯ + 2 + 4]
= (π₯ + 1)(π₯2
β 4π₯ + 7)
Answer.32. (π + π)π
+ (π β π)π
.(π₯ + 1)3
+ (π₯ β 1)3
= [(π₯ + 1) + (π₯ β 1)][(π₯ + 1)2
β (π₯ + 1)(π₯ β 1) + (π₯ β 1)2]
= (2π₯)[(π₯2
+ 2π₯ + 1) β π₯2
+ 1 + (π₯2
β 2π₯ + 1)]
= 2π₯[π₯2
+ 2π₯ + 1 β π₯2
+ 1 + π₯2
+ 1 β 2π₯]
= 2π₯(π₯2
+ 3)
Answer.33. (ππ + π)π
+ (π β π)π
.(2π + 1)3
+ (π β 1)3
= [(2π + 1) + (π β 1)][(2π + 1)2
β (2π + 1)(π β 1) + (π β 1)2]
= (3π)[(4π2
+ 4π + 1) β 2π2
+ 2π β π + 1 + (π2
+ 1 β 2π)]
= 3π[4π2
+ 4π + 1 β 2π2
+ π + 1 + π2
+ 1 β 2π]
= 3π(3π2
+ 3π + 3)
= 9π(π2
+ π + 1)
Answer.34. π(π + π)π
β ππ(π β π)π
8(π₯ + π¦)3
β 27(π₯ β π¦)3
= {2(π₯ + π¦)}3
β {3(π₯ β π¦)}3
= {2(π₯ + π¦) β 3(π₯ β π¦)}[{2(π₯ + π¦)}2
+ {2(π₯ + π¦)}{3(π₯ β π¦)} +
{3(π₯ β π¦)}2
]
= [2π₯ + 2π¦ β 3π₯ + 3π¦][{4(π₯2
+ 2π₯π¦ + π¦2)} + 6(π₯2
β π¦2) +
{9(π₯2
β 2π₯π¦ + π¦2
)}]
= (βπ₯ + 5π¦)[4π₯2
+ 8π₯π¦ + 4π¦2
+ 6π₯2
β 6π¦2
+ 9π₯2
β 18π₯π¦ + 9π¦2
]
= (βπ₯ + 5π¦)(19π₯2
β 10π₯π¦ + 7π¦2)
- 28. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.35. (π + π)π
+ (π β π)π
.(π₯ + 2)3
+ (π₯ β 2)3
= [(π₯ + 2) + (π₯ β 2)][(π₯ + 2)2
β (π₯ + 2)(π₯ β 2) + (π₯ β 2)2]
= (2π₯)[(π₯2
+ 4π₯ + 4) β π₯2
+ 4 + (π₯2
β 4π₯ + 4)]
= 2π₯[π₯2
+ 4π₯ + 4 β π₯2
+ 4 + π₯2
β 4π₯ + 4]
= 2π₯(π₯2
+ 12)
Answer.36. (π + π)π
β (π β π)π
.(π₯ + 2)3
β (π₯ β 2)3
= [(π₯ + 2) β (π₯ β 2)][(π₯ + 2)2
+ (π₯ + 2)(π₯ β 2) + (π₯ β 2)2]
= (4)[(π₯2
+ 4π₯ + 4) + π₯2
β 4 + (π₯2
β 4π₯ + 4)]
= 4[π₯2
+ 4π₯ + 4 + π₯2
β 4 + π₯2
β 4π₯ + 4]
= 4(3π₯2
+ 4)
Answer.37. Prove that :
π.ππΓπ.ππΓπ.ππ+π.ππΓπ.ππΓπ.ππ
π.ππΓπ.ππβπ.ππΓπ.ππ+π.ππΓπ.ππ
= π
L.H.S.β
0.85Γ0.85Γ0.85+0.15Γ0.15Γ0.15
0.85Γ0.85β0.85Γ0.15+0.15Γ0.15
=
(0.85)3+(0.15)3
0.85Γ0.85β0.85Γ0.15+0.15Γ0.15
=
(0.85+0.15){(0.85)2β0.85Γ0.15+(0.15)2
(0.85)2β0.85Γ0.15+(0.15)2
= (0.85 + 0.15)
= 1
= R.H.S
Hππππ, πΏ. π». π = π
. π». π
Answer.38. Prove that :
ππΓππΓππβπΓπΓπ
ππΓππ+ππΓπ+πΓπ
= ππ
L.H.S.β
59Γ59Γ59β9Γ9Γ9
59Γ59+59Γ9+9Γ9
=
(59)3
β(9)3
59Γ59+59Γ9+9Γ9
=
(59β9){(59)2+59Γ9+(9)2}
59Γ59+59Γ9+9Γ9
= (59 β 9)
= 50
= R.H.S
Hππππ, πΏ. π». π = π
. π». π
- 29. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
EXERCISE β 3G
Formula Used : (ππ
+ ππ
+ ππ
β ππππ) = (π + π + π)(ππ
+ ππ
+ ππ
β ππ β ππ β ππ)
Answer.1. (π + π β π)(ππ
+ ππ
+ ππ
β ππ + ππ + ππ)
.(π₯ + π¦ β π§)(π₯2
+ π¦2
+ π§2
β π₯π¦ + π¦π§ + π§π₯) = {π₯ + π¦ + (βπ§)}{π₯2
+ π¦2
+ (βπ§)2
β π₯π¦ β π¦(βπ§) β (βπ§)π₯}
= {π₯3
+ π¦3
+ (βπ§)3
β 3π₯π¦(βπ§)}
= (π₯3
+ π¦3
β π§3
+ 3π₯π¦π§)
Answer.2. (π β π β π)(ππ
+ ππ
+ ππ
+ ππ β ππ + ππ)
.(π₯ β π¦ β π§)(π₯2
+ π¦2
+ π§2
+ π₯π¦ β π¦π§ + π§π₯) = {π₯ + (βπ¦) + (βπ§)}{π₯2
+ (βπ¦)2
+ (βπ§)2
β π₯(βπ¦) β (βπ¦)(βπ§) β
(βπ§)π₯}
= {π₯3
+ (βπ¦)3
+ (βπ§)3
β 3π₯(βπ¦)(βπ§)}
= (π₯3
β π¦3
β π§3
β 3π₯π¦π§)
Answer.3. (π β ππ + π)(ππ
+ πππ
+ πππ + ππ β ππ + π)
.(π₯ β 2π¦ + 3)(π₯2
+ 4π¦2
+ 9 + 2π₯π¦ + 6π¦ β 3π₯) = {π₯ + (β2π¦) + 3}{π₯2
+ (β2π¦)2
+ 32
β π₯(β2π¦) β
(β2π¦)(3) β 3(π₯)}
= {π₯3
+ (β2π¦)3
+ (3)3
β 3π₯(β2π¦)(3)}
= (π₯3
β 8π¦3
+ 27 + 18π₯π¦)
Answer.4. (ππ β ππ + π)(πππ
+ ππππ
+ ππππ + πππ β πππ + ππ)
.(3π₯ β 5π¦ + 4)(9π₯2
+ 25π¦2
+ 15π₯π¦ + 20π¦ β 12π₯ + 16) = {3π₯ + (β5π¦) + 4}{(3π₯)2
+ (β5π¦)2
+ 42
β 3π₯(β5π¦) β
(β5π¦)(4) β 4(3π₯)}
= {(3π₯)3
+ (β5π¦)3
+ (4)3
β 3(3π₯)(β5π¦)(4)}
= (27π₯3
β 125π¦3
+ 64 + 180π₯π¦)
Answer.5. πππππ
+ ππ
+ ππππ
β πππππ
125π3
+ π3
+ 64π3
β 60πππ = (5π)3
+ (π)3
+ (4π)3
β 3 Γ (5π) Γ (π) Γ (4π)
= (5π + π + 4π)[(5π)2
+ (π)2
+ (4π)2
β (5π)(π) β (π)(4π) β (4π)(5π)]
= (5π + π + 4π)(25π2
+ π2
+ 16π2
β 5ππ β 4ππ β 20ππ)
Answer.6. ππ
+ πππ
+ ππππ
β πππππ
π3
+ 8π3
+ 64π3
β 24πππ = (π)3
+ (2π)3
+ (4π)3
β 3 Γ (π) Γ (2π) Γ (4π)
= (π + 2π + 4π)[(π)2
+ (2π)2
+ (4π)2
β (π)(2π) β (2π)(4π) β (4π)(π)]
= (π + 2π + 4π)(π2
+ 4π2
+ 16π2
β 2ππ β 8ππ β 4ππ)
Answer.7. π + ππ
+ πππ
β ππππ
1 + π3
+ 8π3
β 6πππ = (1)3
+ (π)3
+ (2π)3
β 3 Γ (1) Γ (π) Γ (2π)
= (1 + π + 2π)[(1)2
+ (π)2
+ (2π)2
β (1)(π) β (π)(2π) β (2π)(1)]
= (1 + π + 2π)(1 + π2
+ 4π2
β π β 2π β 2π)
Answer.8. πππ + ππππ
+ πππ
β ππππππ
216 + 27π3
+ 8π3
β 108πππ = (6)3
+ (3π)3
+ (2π)3
β 3 Γ (6) Γ (3π) Γ (2π)
= (6 + 3π + 2π)[(6)2
+ (3π)2
+ (2π)2
β (6)(3π) β (3π)(2π) β (2π)(6)]
= (6 + 3π + 2π)(36 + 9π2
+ 4π2
β 18π β 6ππ β 12π)
Answer.9. ππππ
β ππ
+ πππ
+ πππππ
27π3
β π3
+ 8π3
+ 18πππ = (3π)3
+ (βπ)3
+ (2π)3
β 3 Γ (3π) Γ (βπ) Γ (2π)
= (3π β π + 2π)[(3π)2
+ (βπ)2
+ (2π)2
β (3π)(βπ) β (βπ)(2π) β (2π)(3π)]
= (3π β π + 2π)(9π2
+ π2
+ 4π2
+ 3ππ + 2ππ β 6ππ)
- 30. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.10. πππ
+ πππππ
β ππππ
+ ππππππ
8π3
+ 125π3
β 64π3
+ 120πππ = (2π)3
+ (5π)3
+ (β4π)3
β 3 Γ (2π) Γ (5π) Γ (β4π)
= (2π + 5π β 4π)[(2π)2
+ (5π)2
+ (β4π)2
β (2π)(5π) β (5π)(β4π) β (β4π)(2π)]
= (2π + 5π β 4π)(4π2
+ 25π2
+ 16π2
β 10ππ + 20ππ + 8ππ)
Answer.11. π β ππππ
β πππππ
β ππππππ
8 β 27π3
β 343π3
β 126πππ = (2)3
+ (β3π)3
+ (β7π)3
β 3 Γ (2) Γ (β3π) Γ (β7π)
= (2 β 3π β 7π)[(2)2
+ (β3π)2
+ (β7π)2
β (2)(β3π) β (β3π)(β7π) β (β7π)(2)]
= (2 β 3π β 7π)(4 + 9π2
+ 49π2
+ 6π β 21ππ + 14π)
Answer.12. πππ β πππ
β ππππ
β ππππ
125 β 8π₯3
β 27π¦3
β 90π₯π¦ = (5)3
+ (β2π₯)3
+ (β3π¦)3
β 3 Γ (5) Γ (β2π₯) Γ (β3π¦)
= (5 β 2π₯ β 3π¦)[(5)2
+ (β2π₯)2
+ (β3π¦)2
β (5)(β2π₯) β (β2π₯)(β3π¦) β (β3π¦)(5)]
= (5 β 2π₯ β 3π¦)(25 + 4π₯2
+ 9π¦2
+ 10π₯ β 6π₯π¦ + 15π¦)
Answer.13. πβπππ
+ ππβπππ
+ ππ
β πππππ
2β2π3
+ 16β2π3
+ π3
β 12πππ = (β2π)
3
+ (2β2π)
3
+ (π)3
β 3(β2π)(2β2π)(π)
= (β2π + 2β2π + π) [(β2π)
2
+ (2β2π)
2
+ (π)3
β (β2π)(2β2π) β (2β2π)π β π(β2π)]
= (β2π + 2β2π + π)(2π2
+ 8π2
+ π2
β 4ππ β 2β2ππ β β2ππ)
Answer.14. ππππ
β ππ
β ππ
β ππππ
27π₯3
β π¦3
β π§3
β 9π₯π¦π§ = (3π₯)3
+ (βπ¦)3
+ (βπ§)3
β 3 Γ (3π₯) Γ (βπ¦) Γ (βπ§)
= (3π₯ β π¦ β π§)[(3π₯)2
+ (βπ¦)2
+ (βπ§)2
β (3π₯)(βπ¦) β (βπ¦)(βπ§) β (βπ§)(3π₯)]
= (3π₯ β π¦ β π§)(9π₯2
+ π¦2
+ π§2
+ 3π₯π¦ β π₯π¦ + 3π§π₯)
Answer.15. πβπππ
+ πβπππ
+ ππ
β πβππππ
2β2π3
+ 3β3π3
+ π3
β 3β6πππ = (β2π)
3
+ (β3π)
3
+ (π)3
β 3(β2π)(β3π)(π)
= (β2π + β3π + π) [(β2π)
2
+ (β3π)
2
+ (π)3
β (β2π)(β3π) β (β3π)π β π(β2π)]
= (β2π + β3π + π)(2π2
+ 9π2
+ π2
β β6ππ β β3ππ β β2ππ)
Answer.16. πβπππ
β ππ
β πβπππ
β πβπππππ
3β3π3
β π3
β 5β5π3
β 3β15πππ = (β3π)
3
+ (βπ)3
+ (ββ5π)
3
β 3(β3π)(βπ)(ββ5π)
= (β3π β π β β5π) [(β3π)
2
+ (βπ)2
+ (ββ5π)
3
β (β3π)(βπ) β (βπ)(ββ5π) β (ββ5π)(β3π)]
= (β3π β π β β5π)(3π2
+ π2
+ 5π2
+ β3ππ β β5ππ + β15ππ)
Answer.17. (π β π)π
+ (π β π)π
+ (π β π)π
πΏππ‘(π β π) = π₯, (π β π) = π¦ πππ (π β π) = π§,
(π β π)3
+ (π β π)3
+ (π β π)3
= π₯3
+ π¦3
+ π§3
, π€βπππ (π₯ + π¦ + π§) = (π β π) + (π β π) + (π β π) = 0
= 3π₯π¦π§ [β΅ (π₯ + π¦ + π§) = 0 β (π₯3
+ π¦3
+ π§3
) = 3π₯π¦π§]
= 3(π β π)(π β π)(π β π)
Answer.18. (π β ππ)π
+ (ππ β π)π
+ (π β π)π
πΏππ‘(π β 3π) = π₯, (3π β π) = π¦ πππ (π β π) = π§,
(π β 3π)3
+ (3π β π)3
+ (π β π)3
= π₯3
+ π¦3
+ π§3
, π€βπππ (π₯ + π¦ + π§) = (π β 3π) + (3π β π) + (π β π) = 0
= 3π₯π¦π§ [β΅ (π₯ + π¦ + π§) = 0 β (π₯3
+ π¦3
+ π§3) = 3π₯π¦π§]
= 3(π β 3π)(3π β π)(π β π)
- 31. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.19. (ππ β ππ)π
+ (ππ β ππ)π
+ (ππ β ππ)π
πΏππ‘(3π β 2π) = π₯, (2π β 5π) = π¦ πππ (5π β 3π) = π§,
(3π β 2π)3
+ (2π β 5π)3
+ (5π β 3π)3
= π₯3
+ π¦3
+ π§3
, π€βπππ (π₯ + π¦ + π§) = (3π β 2π) + (2π β 5π) + (5π β 3π) = 0
= 3π₯π¦π§ [β΅ (π₯ + π¦ + π§) = 0 β (π₯3
+ π¦3
+ π§3) = 3π₯π¦π§]
= 3(3π β 2π)(2π β 5π)(5π β 3π)
Answer.20. (ππ β ππ)π
+ (ππ β ππ)π
+ (ππ β ππ)π
πΏππ‘(5π β 7π) = π₯, (7π β 9π) = π¦ πππ (9π β 5π) = π§,
(5π β 7π)3
+ (7π β 9π)3
+ (9π β 5π)3
= π₯3
+ π¦3
+ π§3
, π€βπππ (π₯ + π¦ + π§) = (5π β 7π) + (7π β 9π) + (9π β 5π) = 0
= 3π₯π¦π§ [β΅ (π₯ + π¦ + π§) = 0 β (π₯3
+ π¦3
+ π§3) = 3π₯π¦π§]
= 3(5π β 7π)(7π β 9π)(9π β 5π)
Answer.21. ππ(π β π)π
+ ππ(π β π)π
+ ππ(π β π)π
πΏππ‘π(π β π) = π₯, π(π β π) = π¦ πππ π(π β π) = π§,
[π(π β π)]3
+ [π(π β π)]3
+ [π(π β π)]3
= π₯3
+ π¦3
+ π§3
, π€βπππ (π₯ + π¦ + π§) = π(π β π) + π(π β π) + π(π β π) = 0
= 3π₯π¦π§ [β΅ (π₯ + π¦ + π§) = 0 β (π₯3
+ π¦3
+ π§3
) = 3π₯π¦π§]
= 3πππ(π β π)(π β π)(π β π)
Answer.22.
(i) (βππ)π
+ ππ
+ ππ
πΏππ‘ π₯ = (β12), π¦ = 7 πππ π§ = 5
(π₯ + π¦ + π§) = 0
β (π₯3
+ π¦3
+ π§3
) = 3π₯π¦π§
β (β12)3
+ 73
+ 53
= 3 Γ (β12) Γ (7) Γ (5)
β (β12)3
+ 73
+ 53
= β108
(ii) (ππ)π
+ (βππ)π
+ (βππ)π
πΏππ‘ π₯ = 28 , π¦ = (β15) πππ π§ = (β13)
(π₯ + π¦ + π§) = 0
β (π₯3
+ π¦3
+ π§3
) = 3π₯π¦π§
β (28)3
+ (β15)3
+ (β13)3
= 3 Γ (28) Γ (β15) Γ (β13)
β (28)3
+ (β15)3
+ (β13)3
= 16380
Answer.23. (π + π + π)π
β ππ
β ππ
β ππ
= π(π + π)(π + π)(π + π)
πΏ. π». π β (π + π + π)3
β π3
β π3
β π3
= [(π + π) + π]3
β π3
β π3
β π3
= (π + π)3 + π3 + 3(π + π)π Γ [(π + π) + π] β π3 β π3
β π3
= π3 + π3
+ 3ππ(π + π) + π3 + 3(π + π)π Γ [(π + π) + π] β π3 β π3
β π3
= 3ππ(π + π) + 3(π + π)π Γ [(π + π) + π]
= 3(π + π)[ππ + π(π + π) + π2]
= 3(π + π)[ππ + ππ + ππ + π2]
= 3(π + b)[π(π + π) + π(π + π)]
= 3(π + π)(π + π)(π + π)
= π
. π». π
Answer.24. πΊππ£ππ, π + π + π = 0 πππ π, π , π πππ πππ πππ β π§πππ
π2
ππ
+
π2
ππ
+
π2
ππ
= 3
πΏ. π». π β
(π3
+ π3
+ π3)
πππ
- 32. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
β
3πππ
πππ
[β΅ (π + π + π) = 0 β (π3
+ π3
+ π3
) = 3πππ]
β 3
β π
. π». π
Answer.25. πΊππ£ππ π + π + π = 9 πππ π2
+ π2
+ π2
= 35
(π + π + π) = 9 β (π + π + π)2
= 81
β (π2
+ π2
+ π2) + 2(ππ + ππ + ππ) = 81
β 35 + 2(ππ + ππ + ππ) = 81
β (ππ + ππ + ππ) = 23
βΈ« (π3
+ π3
+ π3
β 3πππ) = (π + π + π)(π2
+ π2
+ π2
β ππ β ππ β ππ)
= (π + π + π)[(π2
+ π2
+ π2
) β (ππ + ππ + ππ)]
= 9 Γ (35 β 23)
= 9 Γ 12
= 108
- 33. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
MULTIPLE CHOICE QUESTION (MCQ)
Answer.1. πΏππ‘ π(π₯) = 2π₯2
+ ππ₯
πππππ, (π₯ + 1) ππ π ππππ‘ππ
f(β1) = 0
β2(β1)2
+ π(β1) = 0
β 2 β π = 0
β π = 2
πͺππππππ πΆπππππ βΆ (π)
Answer.2. (249)2
β(248)2
= (249 β 248)(249 + 248)
= (1)(497)
= 497
πͺππππππ πΆπππππ βΆ (π
)
Answer.3.
π₯
π¦
+
π¦
π₯
= β1 β π₯2
+ π¦2
= βπ₯π¦
β π₯2
+ π¦2
+ π₯π¦ = 0
(π₯3
β π¦3
) = (π₯ β π¦)(π₯2
+ π¦2
+ π₯π¦)
= (π₯ β π¦) Γ 0
= 0
πͺππππππ πΆπππππ βΆ (π)
Answer.4. π3
+ π3
+ π3
β 3πππ = (π + π + π)(π2
+ π2
+ π2
β ππ β ππ β ππ)
= 0 Γ (π2
+ π2
+ π2
β ππ β ππ β ππ) [β΅ π + π + π = 0]
= 0
π3
+ π3
+ π3
β 3πππ = 0
βΈ« π3
+ π3
+ π3
= 3πππ
πͺππππππ πΆπππππ βΆ (π
)
Answer.5. (3π₯ +
1
2
) (3π₯ β
1
2
) = 9π₯2
β π
{(3π₯)2
β (
1
2
)
2
} = 9π₯2
β π
(9π₯2
β
1
4
) = 9π₯2
β π
ππ, π =
1
4
πͺππππππ πΆπππππ βΆ (π)
Answer.6. (π₯ + 3)3
= π₯3
+ 33
+ 3 Γ (π₯2) Γ 3 + 3 Γ π₯ Γ (32
)
= π₯3
+ 27 + 9π₯2
+ 27π₯
πΆππππππππππ‘ ππ π₯ = 2
πͺππππππ πΆπππππ βΆ (π
)
Answer.7. (π₯ + π¦)3
β (π₯3
+ π¦3)
= (π₯ + π¦)3
β {(π₯ + π¦)(π₯2
β π₯π¦ + π¦2
)}
= (π₯ + π¦){(π₯ + π¦)2
β (π₯2
β π₯π¦ + π¦2
)}
= (π₯ + π¦){π₯2
+ π¦2
+ 2π₯π¦ β π₯2
+ π₯π¦ β π¦2}
= (π₯ + π¦)(3π₯π¦)
ππ, 3π₯π¦ ππ π ππππ‘ππ.
πͺππππππ πΆπππππ βΆ (π
)
- 34. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
Answer.8. (25π₯2
β 1) + (1 + 5π₯)2
= 25π₯2
β 1 + 1 + 25π₯2
+ 10π₯
= 50π₯2
+ 10π₯
= 10π₯(5π₯ + 1)
ππ, 10π₯ ππ π ππππ‘ππ.
πͺππππππ πΆπππππ βΆ (π
)
Answer.9. π(π₯) = π₯3
β 20π₯ + 5π
πππππ, (π₯ + 5) ππ π ππππ‘ππ.
π(β5) = 0
β(β5)3
β 20(β5) + 5π = 0
ββ125 + 100 + 5π = 0
β5π = 25
βπ = 5
πͺππππππ πΆπππππ βΆ (π)
Answer.10. πΏππ‘ π(π₯) = π₯3
+ 10π₯2
+ ππ₯ + π
(π₯ + 2) = 0 β π₯ = β2
(π₯ β 1) = 0 β π₯ = 1
πππ€, π(β2) = 0 πππ π(1) = 0 [β΅ (π₯ + 2) πππ (π₯ β 1) πππ ππππ‘πππ .]
ππ, π(β2) = 0
β (β2)3
+ 10 Γ (β2)2
+ (β2)π + π = 0
β β8 + 40 β 2π + π = 0
β 2π β π = 32 β¦ (π)
π΄ππ, π(1) = 0
β(1)3
+ 10 Γ (1)2
+ π + π = 0
β1 + 11 + π + π = 0
βπ + π = β11 β¦ (ππ)
π΄πππππ (π) πππ (ππ),
β2π β π + π + π = 32 β 11
β3π = 21
βπ = 7 β¦ (πππ)
πΌππππ (ππ) πππ
(πππ),
β7 + π = β11
βπ = β18
β΄ π = 7 πππ π = β18
πͺππππππ πΆπππππ βΆ (π)
Answer.11. (πππ Γ ππ) = ?
(104 Γ 96) = (100 + 4) Γ (100 β 4)
= (100)2
β (4)2
= 10000 β 16
= 9984
πͺππππππ πΆπππππ βΆ (π)
Answer.12. (πππ Γ πππ) = ?
(305 Γ 308) = 305 Γ (300 + 8)
= (305 Γ 300) + (305 Γ 8)
= (91500 + 2440)
= 93940
πͺππππππ πΆπππππ βΆ (π)
Answer.13. (πππ Γ πππ) = ?
(207 Γ 193) = 207 Γ (200 β 7)
- 35. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (207 Γ 200) β (207 Γ 7)
= (41400 β 1449)
= 39951
πͺππππππ πΆπππππ βΆ (π)
Answer.14. πππ
+ ππ
+ πππ + ππ + ππ + π
4π2
+ π2
+ 4ππ + 8π + 4π + 4 = (2π)2
+ (π)2
+ (2)2
+ 2(2π)(π) + 2(π)(2) + 2(2)(2π)
= (2π + π + 2)2
πͺππππππ πΆπππππ βΆ (π)
Answer.15. (ππ
β ππ β ππ) = ?
(π₯2
β 4π₯ β 21) = π₯2
β 7π₯ + 3π₯ β 21
= π₯(π₯ β 7) + 3(π₯ β 7)
= (π₯ β 7)(π₯ + 3)
πͺππππππ πΆπππππ βΆ (π)
Answer.16. (πππ
+ ππ β π) = ?
(4π₯2
+ 4π₯ β 3) = 4π₯2
+ 6π₯ β 2π₯ β 3
= 2π₯(2π₯ + 3) β 1(2π₯ + 3)
= (2π₯ + 3)(2π₯ β 1)
πͺππππππ πΆπππππ βΆ (π)
Answer.17. πππ
+ πππ + π = ?
6π₯2
+ 17π₯ + 5 = 6π₯2
+ 15π₯ + 2π₯ + 5
= 3π₯(2π₯ + 5) + 1(2π₯ + 5)
= (2π₯ + 5)(3π₯ + 1)
πͺππππππ πΆπππππ βΆ (π)
Answer.18. πΏππ‘ π₯3
+ 2π₯2
β π₯ β 2 = π(π₯)
π(β1) = {(β1)3
+ 2(β1)2
β (β1) β 2}
= (β1 + 2 + 1 β 2)
= 0
πͺππππππ πΆπππππ βΆ (π)
Answer.19. πππ
+ πππ
+ ππ + π = ?
3π₯3
+ 2π₯2
+ 3π₯ + 2 = π₯2(3π₯ + 2) + 1(3π₯ + 2)
= (3π₯ + 2)(π₯2
+ 1)
πͺππππππ πΆπππππ βΆ (π
)
Answer.20. π + π + π = 0
βπ3
+ π3
+ π3
β 3πππ = 0
βπ3
+ π3
+ π3
= 3πππ
(
π2
ππ
+
π2
ππ
+
π2
ππ
) =
1
πππ
(π3
+ π3
+ π3)
=
1
πππ
(3πππ)
= 3
πͺππππππ πΆπππππ βΆ (π
)
Answer.21. (π₯3
+ π¦3
+ π§3
β 3π₯π¦π§) = (π₯ + π¦ + π§)(π₯2
+ π¦2
+ π§2
β π₯π¦ β π¦π§ β π§π₯)
= (π₯ + π¦ + π§)(π₯2
+ π¦2
+ π§2
+ 2π₯π¦ + 2π¦π₯ + 2π§π₯ β 3π₯π¦ β 3π¦π§ β 3π§π₯)
- 36. CLASS IX WWW.Vedantu.com RS Aggarwal solutions
= (π₯ + π¦ + π§)[(π₯2
+ π¦2
+ π§2
+ 2π₯π¦ + 2π¦π₯ + 2π§π₯) β 3(π₯π¦ + π¦π§ + π§π₯)]
= (π₯ + π¦ + π§)[(π₯ + π¦ + π§)2
β 3(π₯π¦ + π¦π§ + π§π₯)]
= 9 Γ [(9)2
β 3 Γ (23)]
= 9 Γ (81 β 69)
= 9 Γ 12
= 108
πͺππππππ πΆπππππ βΆ (π)
Answer.22. (
π
π
+
π
π
) = β1
β
π
π
+
π
π
= β1
β
π2+π2
ππ
= β1
βπ2
+ π2
+ ππ = 0
(π3
β π3) = (π β π)(π2
+ ππ + π2
)
= (π β π) Γ 0
= 0
πͺππππππ πΆπππππ βΆ (π
)